Indian Institute of Management Indore is conducting IPMAT on Friday,16 July 2021. This post takes you through a detailed analysis of the exam. Go through the IPMAT Exam Analysis and Solutions to previous year papers to understand the difficulty level of the examination and to enhance your preparation.

The examination generally consists of a total of 400 marks. Candidates can go through the section-wise IPMAT Analysis to improve their preparation for the upcoming IIM IPM Exam 2021.

Candidates can learn about the exam's difficulty level, the number of good attempts, and the questions asked from this post.

IPMAT Exam Analysis Trend

The following points explain the previous five-year exam analysis of the IIM IPM Exam regarding the type of questions asked, mode of exam, sections involved, and marking scheme. This can help you improve your preparation and score higher.

Type of questions:- Multiple Choice Questions MCQs and Short Answer (SA) questions.

Sections:- generally three sections, two of quantitative ability and one of verbal ability.

Maximum Marks:- The examination generally consists of a total of 400 marks.

Marking Scheme:- 4 marks are awarded for every correct answer, 1 mark is deducted for every wrong answer in MCQs and there is no negative marking in Short answers.

Sectional cut-off applied in all three sections.

The overall difficulty level of IPMAT, considering previous years’ examination, is moderate to difficult.

Verbal Ability is the most scoring section in the examination with about 70-80% of questions from easy to moderate difficulty.

Quantitative ability MCQ is generally the most difficult section with about 40-50% of questions from moderate to hard difficulty.

Question 1: In a division problem, the product of quotients and the remainder is 24 while their sum is 10. If the divisor is 5 then the dividend is __________. (IPMAT 2020)

Solution 1 :

In this problem, we have given that the product of quotient(q) and the remainder(r) is 24. i.e q x r = 24 -( i )

And it is also given that their sum is 10 i.e q + r = 10 -( ii )

Divisor(d) = 5.

( i ) and ( ii ) represent the sum and product of a quadratic equation, from ( ii ) putting r = q - 10, in ( i ), we will get q x (10 - q) = 24, i.e q^{2}-10q + 24 = 0,

Which will give q = 6 or 4.

Therefore one of them will be 6 and the other one will be 4. We do not know yet whether Quotient is 6 and Remainder is 4 or vice versa. We know that Dividend(n) = divisor(d) x quotient(q) + remainder(r) .

From the given data, n = 5 x q + r. - ( iii ),

As the divisor is 5 we infer that the remainder obviously will be smaller than 5, therefore the remainder will be 4 and the quotient will be 6. Putting the above-decided value in ( iii ). n = 5(6) + 4 , therefore n = 34.

Question 2 : The shortest distance from the point (-4,3) to the circle x2 + y2 = 1 is __________. ( IPMAT 2020 )

Solution 2.

In this question, we have been given the equation of a circle and the coordinates of a point.

From the standard equation of circle i.e x2 + y2 = a2, we have a = 1.

The shortest distance of the point P(-4, 3) will be the distance from point P to the point where a straight line drawn from P to the origin of the circle intersects the circle. As shown in the figure below the shortest distance is PM.

Question 3 : The value of

Question 4: The minimum value of f(x)=|3-x|+|2+x|+|5-x| is equal to __________.

Solution.

Form f(x)=|3-x|+|2+x|+|5-x| We know that critical points would be x = 3 , x = -2 , and x = 5.

The slope of the line changes at these critical points.

Example

Now we need to select the critical point which has the lowest value of y axis, from the figure above f(3) will have the lowest value.

f(3) = |3 - 3|+|2 + 3|+| 5 - 3|

f(3) = 0 + 5 + 2 .

f(3) = 7.

Hence the answer is 7.

Question 5: Ashok purchased pens and pencils in the ratio 2: 3 during his first visit and paid Rs. 86 to the shopkeeper. During his second visit, he purchased pens and pencils in the ratio of 4: 1 and paid Rs. 112. The cost of a pen, as well as a pencil in rupees, is a positive integer. If Ashok purchased four pens during his second visit, then the amount he paid in rupees for the pens during the second visit is __________. (IPMAT 2020)

Solution.

We have been given that when the person spends Rs. 86 the ratio of pen and pencil is 2:3.

And when he pays Rs. 112 the ratio of pen and pencil is 4:1. And in this visit, the total number of pens is 4.

Let x be the price per pen and y be the price per pencil.

Therefore we have

4x + y = 112

2x + 3y = 86 (From first visit)

It is in Ratios, so it can be 4x + 6y = 86 (or)

6x + 9y = 86 (or)

8x + 12y = 86 and so on.

As we know 4x + y = 112 we can infer that 4x + 6y = 86 is not possible and hence all the equations after this are not possible.

So now we have 4x + y = 112

2x + 3y = 86

Solving the above equation we will get y = 12.

Therefore 4x + 12 = 112

=> 4x = cost of pens = 112 -12 = 100

Hence the answer is 100.

Question 6: In a four-digit number, the product of thousands digit and units digit is zero while their difference is 7. The product of the middle digits is 18. The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit. The four-digit number is __________.

Solution.

Let the four-digit number be ABCD. D be the unit place number, C is the tens digit number, B be the hundreds digit number and A be the thousand digit number.

Given data

A x D = 0

This implies one of them will be zero, as A is the rightmost place its value cannot be zero.

Therefore D = 0.

A - D = 7

As D = 0, A will be 7.

B x C = 18

B and C can hold values from 0-9 only, looking at the possible combinations BC can be 92( or 29) or 63(or 36).

Given that the difference of AD is the same as the difference of BC therefore B is 9 and C is 2.

Hence the answer is 7920.

Question 7: Two friends run a 3-kilometer race along a circular course of length 300 meters. If their speeds are in the ratio 3:2, the number of times the winner passes the other is __________.

Solution.

Let the speed of the fast one be 3x and the speed of the slow one be 2x.

Time required for the fast one to cover one complete round = tf = 300/3x

Time required for the slow one to cover one complete round = ts = 300/2x

For finding out the time at which both meet for the first time we need to find lcm of tf and ts

Tmeet = 300/x (lcm of tf and ts)

Distance traveled by the fast one when they meet = Tmeet * speed of fast one

= 300/x * 3x

= 900 m

Therefore they will first meet when the fast one has covered 900m, then they will meet at 1800, and then for the last at 2700 meters.

Therefore the winner will pass the slow one exactly 3 times.

IPMAT Quantitative Ability MCQ Important questions

Question 1: The probability that a randomly chosen factor of 1019 is a multiple of 1015 is

1/25

1/12

1/20

1/16

Solution.

Given number is 1019 , the factor of the given number are 219 x 519

Therefore the total factors of 1019 are (19 + 1)(19 + 1) = 400.

Multiples of 1015 are 215 x 515, This is of the form 2x × 5y

We need a number that is a multiple of 2x × 5y and also a factor of 219 × 519. Therefore the possible values of x = 15 , 16 , 17 ,18 , 19

And possible values of y = 15 , 16 , 17 ,18 , 19

Therefore there are 5 possible values of x and 5 possible values of y.

Favourable outcome = 5 * 5 = 25

Total number of outcomes = 400

Probability = 25/400 = 1/16.

Hence the correct option is option D.

Question 2: The number of acute-angled triangles whose sides are three consecutive positive integers and whose perimeter is at most 100 is

28

29

31

33

Solution.

We have been given three conditions:-

Triangle sides (a+b > c)

Acute angle triangle (a2 + b2 > c2)

Perimeter <= 100 (a + b + c <= 100)

Let's take random values for three consecutive integers: 1, 2, 3

This selection is not possible as it does not satisfy the first condition that the sum of two sides of a triangle should be greater than the third side. Therefore we rule this out. : 2, 3, 4

The first condition is satisfied, but the second condition ie Acute angle property is Sum of the square of any two sides has to be greater than the square of the third side 22 + 32 > 42

As 13 < 16. (So this condition is also not satisfied) : 3, 4, 5

The first condition is satisfied, but the second condition ie Acute angle property is Sum of the square of any two sides has to be greater than the square of the third side 32 + 42 > 52

As 25 = 25. (So this condition is also not satisfied) : 4, 5, 6

Since all the conditions mentioned above are satisfied, 4,5,6 will be our first consecutive integer set. Then (5,6,7) works and so on.

When the numbers become larger squares become increasingly larger

So after 4,5,6, all sets will satisfy condition 2.

Now we need to know the last possible consecutive number set based on the third condition.

Which gives 32,33,34 -Perimeter is at most 100

Therefore from the first 32 Combinations, the First 3 are ruled out. So in total 29 combinations are possible.

Hence choice B is the correct answer.

Question3 : The value of cos2/8+ cos23/8 + cos25/8 + cos27/8 is.

Question 5: A 2 × 2 matrix is filled with four distinct integers randomly chosen from the set {1,2,3,4,5,6}, Then the probability that the matrix generated in such a way is singular is

2/45

1/45

4/15

1/15

Solution.

We know that the matrix will be singular only if its determinant is zero.

Therefore, the given matrix be

a

c

b

c

= |acdb| = ab - cd = 0, Or ab = cd

Using hit and trial from the given set {1,2,3,4,5,6}.

Let one of a,b,c or d is 5, then ab ≠ cd, because in the given set {1,2,3,4,5,6} there is only one multiple of 5. So now we have to select 4 integers from {1,2,3,4,6}

Let's take 3 or 6, since 6 is a multiple of 3 we will need to pick both of them. Therefore now the equation becomes 3b = 6d

From the above equation we can infer that, b > d

{b, d} ⊂ {1, 2, 4}

If b = 4, d = 2

If b = 2, d = 1

Taking 3 and 6 we observed that only 2 choices were possible i.e {a, b, c, d} = {1, 2, 3, 6} or {a, b, c, d} = {4, 2, 3, 6}. Hence, we can select four numbers from the six in 6C4 = 15 ways

Therefore there will be 15 ways to select them, and 4! ways to arrange them in the matrix.

So, the total number of matrices that can be formed is 15 x 4!.

Once we select {a, b, c, d} = {1, 2, 3, 6} or {a, b, c, d} = {4, 2, 3, 6}, for the matrix to be singular 2,3 or 4,3 should be on the same diagonal. Hence , only ⅓ of the arrangements yield a singular matrix.

Only one-third of the arrangements yield a singular matrix and the others don’t.

Probability of getting a singular matrix = 2( 4! ⅓ )154! = 2/45.

(Q.1 - 6)Read the following passage and choose the answer that is closest to each of the questions that are based on the passage.
Supposing half a dozen or a dozen men were cast ashore from a wreck on an uninhabited island and left to their own resources, one of course, according to his capacity, would be set to one business and one to another; the strongest to dig and to cut wood, and to build huts for the rest: the most dexterous to make shoes out of bark and coats out of skins; the best educated to look for iron or lead in the rocks, and to plan the channels for the irrigation of the fields. But though their labours were thus naturally severed, that small group of shipwrecked men would understand well enough that the speediest progress was to be made by helping each other-not by opposing each other; and they would know that this help could only be properly given so long as they were frank and open in their relations, and the difficulties which each lay under properly explained to the rest. So that any appearance of secrecy or separateness in the actions of any of them would instantly, and justly, be looked upon with suspicion by the rest, as the sign of some selfish or foolish proceeding on the part of the individual. If, for instance, the scientific man were found to have gone out at night, unknown to the rest, to alter the sluices, the others would think, and in all probability rightly think, that he wanted to get the best supply of water to his own field; and if the shoemaker refused to show them where the bark grew which he made the sandals of, they would naturally think, and in all probability rightly think, that he didn't want them to see how much there was of it, and that he meant to ask from them more corn and potatoes in exchange for his sandals than the trouble of making them deserved. And thus, although each man would have a portion of time to himself in which he was allowed to do what he chose without let or inquiry - so long as he was working in that particular business which he had undertaken for the common benefit, any secrecy on his part would be immediately supposed to mean mischief; and would require to be accounted for, or put an end to: and this all the more because, whatever the work might be, certainly there would be difficulties about it which, when once they were well explained, might be more or less done away with by the help of the rest; so that assuredly every one of them would advance with his labour not only more happily, but more profitably and quickly, by having no secrets, and by frankly bestowing, and frankly receiving, such help as lay in his way to get or to give.

Q1. When a dozen men are cast away on an imaginary island, the best educated would look for metals in rocks because

metals can be used to make weapons.

such an island probably has unexploited resources.

he may find it beneath him to dig or cut or make shoes.

He is suited for such work.

Q2. The author states that any appearance of secrecy or separateness would instantly and justly be looked upon with suspicion. From this statement we may infer that

what is secret is not what is separate

secrecy is not exactly the same as separateness

it is natural to be suspicious of secrecy

it only takes an instant for a relationship to deteriorate

Q3. The instance of the shoemaker who refuses to show his source and asks for more corn and potatoes, is an example of

a strong bargain.

unfair practice.

the system of barter.

the intent to make trouble.

Q4. According to the author, whatever one's work might be

hardships are going to be part of it.

one cannot keep complaining.

one should expect others to assure of help and advance our labours.

one must offer help to others in order to receive help.

Q5. The author's belief is that for progress to happen

a team should consist of people with multiple talents.

cooperation among team members is essential.

one must deal with those who are secretive.

transparency among all concerned is mandatory.

Q6. The writer makes a hypothesis, which can be related to

With adequate preparation and strategy, you should be able to clear the IPMAT Exam on the first attempt. Joining a good coaching institute for guidance is recommended but not mandatory. In addition, referring to the appropriate books and study materials will aid in passing the exam with good marks.

How to prepare for the IPMAT exam?

To score good marks, candidates must follow expert preparation tips. Before you begin your preparation, review the complete test pattern and curriculum for the IPMAT Exam and create a comprehensive study plan. Study each topic by referring to the appropriate IPMAT Books, and make sure to finish studying the topics specified for each day without excuses. Participate in the Mock Test Series on a regular basis to assess your preparation and Solve previous year papers to determine the difficulty level of the paper and to estimate the different types of questions asked in the examination.

What is the weightage of each section in the IPMAT 2021 exam?

The question paper is divided into two categories, according to the IPMAT Syllabus: Quantitative Ability (Multiple Choice and Short Answer Questions) and Verbal Ability (Multiple Choice and Short Answer Questions) (Multiple Choice Questions). There are 40 questions in the Quantitative Ability Multiple Choice portion, and 20 questions in the Quantitative Ability Short Answer section. There are 40 multiple-choice questions in the Verbal Ability section.

What are the benefits of understanding IPMAT Exam analysis?

Candidates can understand the exam's difficulty level.

Recognize the successful attempts.

Also, be aware of the types of questions that will be asked in each segment.

What is a good attempt in the IPMAT Exam?

According to the IPMAT Exam Analysis, a good attempt depends upon various factors like difficulty level of exam, number of students appearing, etc. Generally speaking, anything between 65-70 correct attempts would be a good attempt.